Problem 50 - College Algebra and Calculus: An Applied Approach

Question

(a) write the system of linear equations as a matrix equation \(A X=B\), and (b) use Gauss-Jordan elimination on the augmented matrix \([A: B]\) to solve for the matrix \(X\). $$ \left\\{\begin{aligned} x+y-3 z &=-1 \\ -x+2 y &=1 \\ x-y+z &=2 \end{aligned}\right. $$

Step-by-Step Solution

Answer

The solution to the system of equations is: \(x = 1\), \(y = 1.5\), \(z = 1.5\)

Step 1: Convert the system of linear equations into a matrix equation

We can write the system of equations in the form of \(AX = B\), where \(A\) is the coefficient matrix, \(X\) is the matrix containing the variables (\(x, y, z\)) and \(B\) is the matrix of the constants. From the given equation, we find: \( A = \[[1, 1, -3], [-1, 2, 0], [1, -1, 1]\]\), \(X = \[[x], [y], [z]\]\) and \(B = \[[-1], [1], [2]\]\)

Step 2: Create the augmented matrix [A:B]

We now create an augmented matrix by appending \(B\) onto \(A\). This gives: \([A:B] = \[[1,1,-3|-1],[-1,2,0|1],[1,-1,1|2]\]\)

Step 3: Apply Gauss-Jordan Elimination

Starting with our augmented matrix, we perform row operations to make a designated pivot element 1, and every other element in its column 0: \n - Swap row 2 and row 1 to bring the -1 at the beginning to the top: \[[-1, 2, 0 | 1], [1, 1, -3 | -1], [1, -1, 1 | 2]\]\n - Multiply row 1 by -1 for simplicity: \[[1, -2, 0 | -1], [1, 1, -3 | -1], [1, -1, 1 | 2]\]\n - Subtract row 1 from row 2 and row 3 to make first element of these rows equal to zero: \[[1, -2, 0 | -1], [0, 3, -3 | 0], [0, 1, 1 | 3]\]\n - Swap row 2 and row 3 to bring the 1 at the beginning of the second column to the middle: \[[1, -2, 0 | -1], [0, 1, 1 | 3], [0, 3, -3 | 0]\]\n - Subtract 3 times row 2 from row 3 to make the second element of the third row 0: \[[1, -2, 0 | -1], [0, 1, 1 | 3], [0, 0, -6 | -9]\]\n - Divide row 3 by -6 to make the last element of the third row (the last pivot) a 1: \[[1, -2, 0 | -1], [0, 1, 1 | 3], [0, 0, 1 | 1.5]\]\n - Finally, subtract row 3 from row 2 to make the third element of the second row 0 and add 2 times row 2 to row 1 to make the second element of the first row 0, which gives us the final form: \[[1, 0, 0 | 1], [0, 1, 0 | 1.5], [0, 0, 1 | 1.5]\]\n These operations yield a matrix in reduced row echelon form, giving us the solution to the system: \(x = 1\), \(y = 1.5\), \(z = 1.5\)

Key Concepts

Matrix Equation

The concept of a matrix equation is foundational to understanding linear algebra, especially when dealing with systems of linear equations. A matrix equation formulates a system of equations as a single, concise mathematical statement. The equation has the format \(AX = B\), where \(\) represents the matrix containing the coefficients of the variables from the linear system, \(\) is a column matrix of the variables, and \(\) is a column matrix of constants.

For the given system of equations, the coefficient matrix \(\) is a 3x3 matrix, reflecting the three equations and three unknown variables \(\).

• The first column of \(\) contains the coefficients of \(\),
• the second column contains coefficients of \(\), and
• the third column holds coefficients of \(\).

The result is a representation that allows for the efficient application of operations that will solve for \(\), the variables of our system.

System of Linear Equations

A system of linear equations consists of multiple linear equations that share a common set of variables. The goal is to find the values of these variables that satisfy all equations simultaneously. Systems can be classified by the number of solutions they possess—unique, infinite, or none—which depends on the equations' configurations. Solving systems like the aforementioned one typically involves algebraic methods or matrix operations – the latter exemplified by the Gauss-Jordan elimination method.

To express the system as a matrix equation, each linear equation is transformed into a row in our coefficient matrix \(\), with the corresponding constants forming the constant matrix \(\). This prepares the system for matrix operations aimed at finding the solution vector \(\).

Reduced Row Echelon Form

The reduced row echelon form (RREF) is a state of a matrix characterized by certain criteria. In RREF, the leading entry (or pivot) of each row is 1, and it is the only non-zero entry in its column. Moreover, the pivot of each subsequent row is to the right of the pivot above it, and rows containing only zeros are at the bottom of the matrix if there are any.

RREF is the end goal of the Gauss-Jordan elimination process as it makes it straightforward to interpret the solution of a system of linear equations. For instance, the matrix from the exercise culminates in an RREF, which directly illustrates the solutions for \(\), \(\), and \(\). When a matrix achieves this echelon form, solving the corresponding system becomes a matter of reading off the values.

Augmented Matrix

The term augmented matrix is utilized when we combine the coefficient matrix \(\) with the constants matrix \(\) of a system of linear equations to make a new matrix \([A:B]\). This augmented matrix streamlines computations by keeping the coefficients and constants in close proximity during manipulations.

Demonstrated in the step-by-step solution, the Gauss-Jordan elimination procedure is applied to the augmented matrix. The process involves a series of row operations with the ultimate aim of transforming \([A:B]\) to the form where \(\) is the identity matrix, isolating \(\) to display the solution vector directly. This singular matrix, which contains both \(\) and \([B>\), encapsulates the entire system of equations and is manipulated as one entity throughout the elimination process.